Also, note that $24$ has to divide $n^2(n+1)^2(n-1)(n-2)$ since $24$ divides $(n-2)(n-1)n(n+1)$ since it is a product of $4$ consecutive integers.

Further, $2$ divides $n(n+1)$. Hence, we know that at-least $48$ divides $f(n)$. Hence, the desired number should be a multiple of $48$ and must divide $288$.

Can you finish it from here by looking at $f(n)$ for couple of other values?
Move the mouse over the gray area for the answer.

Since the largest number must be a multiple of $48$ and must divide $288$. The possible options are $$\{48, 96, 144, 288\}$$
$$\text{$f(4) = 2400$. And $144,288$ doesn't divide $2400$. Hence, the largest number cannot be $144$ or $288$.}$$
$$\text{$f(5) = 10,800$. And $96$ doesn't divide $10,800$. Hence, the largest number cannot be $96$. Hence, the largest number is $48$.}$$

Factor as $(n-2)(n-1)n^2(n+1)^2$. It is easy to see that no prime greater than $3$ must divide this product. (For if $p$ is a prime greater than $3$, let $n-3=p$.)

Our product is obviously divisible by $3$. By choosing $n-1=3$, we can make sure that $3^2$ does not divide our product.

To minimize the power of $2$ that divides our product, we make $n-2$ (respectively, $n-1$) divisible by $4$ but not by $8$. Then $n^2$ (respectively, $(n+1)^2$) is divisible by $4$ but not by $8$. Thus $2^4$ must divide our product, and a higher power of $2$ need not.

We conclude that the largest positive integer that always divides our product is $(3^1)(2^4)$.